For the reaction : 2AgCl(s) + H_2(g)\ (0.4\ atm) → 2Ag(s) + 2H^+(0.1\ M) + 2Cl^-(0.2\ M) Calculate emf of the cell at 25 °C. Given : Δ G°= -43500 J mol^-1 [log 10 = 1, 1 F = 96500 C mol^-1]

Ecell 0.314 V (with E°cell = 0.225 V)

ChemistryCBSE Class 12 · Board exam 2026, set 56/1/1 · 5 marks · Long answer

CBSE Class 12 Chemistry 2026: question 33(b)

Set in the CBSE Class 12 board exam of 2026, paper 56/1/1, section E, question 33(b), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33(b) · 5 marks · Long answerElectrochemistry

For the reaction :
$2 AgCl (s) + H_{2} (g) (0.4 atm) \to 2 Ag (s) + 2 H^{+} (0.1 M) + 2 C l^{-} (0.2 M)$
Calculate emf of the cell at 25 $^{\circ}$ C.
Given : $Δ G^{\circ} = - 43500$ J mol $^{- 1}$
[log 10 = 1, 1 F = 96500 C mol $^{- 1}$ ]

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 56/1/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Get $E^{\circ}$ from the given $Δ G^{\circ}$ using $Δ G^{\circ} = - n F E^{\circ}$ ; first decide n from the equation.

Hint 2

In Q, leave out the solids, square [H $^{+}$ ] and [Cl $^{-}$ ], and put the pressure of H $_{2}$ in the denominator.

Hint 3

log Q works out to a negative number, so the emf comes out larger than $E^{\circ}$ .

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

From the reaction, 2 electrons are transferred ( $H_{2} \to 2 H^{+} + 2 e^{-}$ ), so $n = 2$ . Use $Δ G^{\circ} = - n F E_{ce l l}^{\circ}$ .
½ mark
$E_{ce l l}^{\circ} = \frac{- Δ G ^{\circ}}{n F} = \frac{43500}{2 \times 96500} = 0.2254 V$
1 mark
Nernst equation at 298 K:
$E_{ce l l} = E_{ce l l}^{\circ} - \frac{0.059}{2} lo g \frac{[ H ^{+} ] ^{2} [ C l ^{-} ] ^{2}}{p _{H_{2}}}$
(AgCl and Ag are solids, so they do not appear in Q.)
½ mark
Calculate the reaction quotient:
$Q = \frac{( 0.1 ) ^{2} ( 0.2 ) ^{2}}{0.4} = \frac{0.01 \times 0.04}{0.4} = 1 0^{- 3}$
1 mark
$E_{ce l l} = 0.2254 - 0.0295 lo g 1 0^{- 3} = 0.2254 - 0.0295 \times (- 3)$
1 mark
$E_{ce l l} = 0.2254 + 0.0885 = 0.3139 V \approx 0.314 V$
1 mark

Answer: $E_{ce l l} \approx 0.314$ V (with $E_{ce l l}^{\circ} = 0.225$ V)

Where the marks are

Identifies n = 2 and uses $Δ G^{\circ} = - n F E^{\circ}$ (½ mark)
Calculates $E_{ce l l}^{\circ} = 0.225$ V (1 mark)
Writes the Nernst equation with the correct reaction quotient (½ mark)
Evaluates Q = $1 0^{- 3}$ (1 mark)
Substitutes into the Nernst equation (1 mark)
Obtains $E_{ce l l} \approx 0.314$ V (1 mark)

Revise the chapter

Chapter: Electrochemistry (notes, formulas, where marks are lost and every past question)
Topic: Nernst Equation, Gibbs Energy and Equilibrium Constant

Practise this chapterBoard-style questions from Electrochemistry, one at a time, marked against the answer key and marking points.Back to the full paper2026 board paper, set 56/1/1: every question, by section.

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CBSE Class 12 Chemistry 2026: question 33(b)

Hints

Step-by-step solution

Revise the chapter

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